Let `A={x in R :-1lt=xlt=1}=B` . Then, the mapping `f: A->B` given by `f(x)=x|x|` is (a) injective but not surjective (b) surjective but not injective (c) bijective (d) none of these

The function f: R->R , f(x)=x^2 is (a) injective but not surjective (b) surjective but not injective (c) injective as well as surjective (d) neither injective nor surjective

The function f: R->R , f(x)=x^2 is (a) injective but not surjective (b) surjective but not injective (c) injective as well as surjective (d) neither injective nor surjective

Let R be the set of real numbers. If f:R->R is a function defined by f(x)=x^2, then f is (a) injective but not surjective (b) surjective but not injective (c) bijective (d) none of these

f: R->R given by f(x)=x+sqrt(x^2) is (a) injective (b) surjective (c) bijective (d) none of these

f: R->R given by f(x)=x+sqrt(x^2) is (a) injective (b) surjective (c) bijective (d) none of these

Let R be the set of real numbers. If f:R->R is a function defined by f(x)=x^2, then f is (a) injective but not surjective (b) surjective but not injective (c) bijective (d) non of these

Let A{x :-1lt=xlt=1}a n df: Avec such that f(x)=x|x|, then f is (a) bijection (b) injective but not surjective (c)Surjective but not injective (d) neither injective nor surjective

Let A}x :-1lt=xlt=1}a n df: Avec A such that f(x)=x|x|, then f is (a) bijection (b) injective but not surjective Surjective but not injective (d) neither injective nor surjective

Let R be the set of real numbers. If f:R->R is a function defined by f(x)=x^2, then f is injective but not surjective surjective but not injective bijective none of these (a) injective but not surjective (b) surjective but not injective (c) bijective (d) non of these

Let A}x:-1<=x<=1} and f:A rarr such that f(x)=x|x|, then f is a bijection (b) injective but not surjective Surjective but not injective (d) neither injective nor surjective